Quartz 5

Fundamental group isomorphisms

3 min read

Proposition

Let be a based map. Then the function

is well defined and satisfies:

  1. is a group homomorphism
  2. based homotopic to
  3. If based maps, then

Proof

Well defined? If as paths, then as paths.

  1. preserves unit composition law respected
  2. relative to relative to
  3. for all
  4. for all

Notation

Proposition

path in induces a group isomorphism

(intuitively, we just go from to , then do the loop and then go back to along the same path) It satisfies:

  1. as paths
  3. If , then
  4. If st and then the following diagram commutes:
\usepackage{tikz-cd}
 
\begin{document}
\begin{tikzcd}
\pi_1(X,x_0) \arrow[r,"f_*"] \arrow[d,"u_{\#}"] & \pi_1(Y,y_0)\arrow[d,"(f\circ u)_{\#}"]\\
\pi_1(X,x_1)\arrow[r, "f_*"] & \pi_1(Y,y_1)
\end{tikzcd}
\end{document}
  1. If , is automorphism of given by conjugation in

Proof

The only interesting part is 4:

Warning

path connected NOT a cannonical isomorphism, it depends on path . NB Abstract properties eg being trivial, abelian, … make sense without specifiying

Lemma

, , . Then define . Then the following diagram commutes:

\usepackage{tikz-cd}
\begin{document}
\begin{tikzcd}
\pi_1(X,x_0) \arrow[r,"f_*"] \arrow[rd, "g_*"] & \pi_1(Y,f(x_0)) \arrow[d, "u_{\#}"]\\
& \pi_1(Y,g(x_0))
\end{tikzcd}
\end{document}

Proof

Idea

For a path in based at we want to construct the following two paths in based at and prove that they are homotopy equivalent.

gx0---g o gamma---->gx0
|                  ^
|                  |
u^-1               u
|                  |
|                  |
V                  |
fx0----f o gamma--->fx0

Note that and We literally put coordinates on this square.

Step 1

with by concatenating: , and Then relative to where

Step 2

Let be a loop based at in . Set . So and and relative to ie as paths from to and so done.

Theorem

homotopy equivalence, Then is an isomorphism

Proof

Let homotopy inverse of . Say . Let , a path in from to . By previous lemma (where functions are from to ) Also is an isomorphism so So the first map is injective, while the second is surjectvie because there is an isomorphism . Reverse roles of and : so is surjective. Now see the commutative diagram (by proposition):

\usepackage{tikz-cd}
\begin{document}
\begin{tikzcd}
\pi_1(X,x_0)\arrow[r,"f_*","inj"'] \arrow[d,"u_{\#}"] & \pi_1(Y,f(x_0)) \arrow[d,"(f\circ u)_{\#}"] \\
\pi_1(X,g\circ f(x_0)) \arrow[r,"f_*", "surj"'] & \pi_1(Y,f\circ g\circ f(x_0))
\end{tikzcd}
\end{document}

But then has to be a bijection and it is already a homomorphism so it is an isomorphism.

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